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The problem asks to find the volume of the solid obtained by rotating the region bounded by the curves \( y = x^2 \) and \( y = 3x \) about the line \( y = -1 \).

### Step-by-Step Solution:

1. **Set up the curves**: 
   - The first curve is \( y = x^2 \), a parabola.
   - The second curve is \( y = 3x \), a straight line.
   
   To find the limits of integration (the points where the curves intersect), set the equations equal to each other:
   \[
   x^2 = 3x
   \]
   Rearrange:
   \[
   x^2 - 3x = 0
   \]
   Factor:
   \[
   x(x - 3) = 0
   \]
   So, the solutions are \( x = 0 \) and \( x = 3 \). Thus, the region is bounded between \( x = 0 \) and \( x = 3 \).

2. **Set up the formula for the volume**:
   Since we're rotating around the line \( y = -1 \), we need to adjust each function by shifting them down by 1 unit (because the axis of rotation is below the curves).
   - For \( y = x^2 \), the distance from the axis of rotation is \( x^2 - (-1) = x^2 + 1 \).
   - For \( y = 3x \), the distance from the axis of rotation is \( 3x - (-1) = 3x + 1 \).
   
   Using the method of washers, the volume formula is:
   \[
   V = \pi \int_{0}^{3} \left[ (3x + 1)^2 - (x^2 + 1)^2 \right] dx
   \]

3. **Expand the terms**:
   - \( (3x + 1)^2 = 9x^2 + 6x + 1 \)
   - \( (x^2 + 1)^2 = x^4 + 2x^2 + 1 \)

   So the integral becomes:
   \[
   V = \pi \int_{0}^{3} \left( (9x^2 + 6x + 1) - (x^4 + 2x^2 + 1) \right) dx
   \]
   Simplify the integrand:
   \[
   V = \pi \int_{0}^{3} \left( -x^4 + 7x^2 + 6x \right) dx
   \]

4. **Integrate**:
   \[
   V = \pi \left[ \frac{-x^5}{5} + \frac{7x^3}{3} + 3x^2 \right]_{0}^{3}
   \]
   Evaluating at \( x = 3 \):
   \[
   V = \pi \left[ \frac{-3^5}{5} + \frac{7 \cdot 3^3}{3} + 3 \cdot 3^2 \right]
   \]
   \[
   V = \pi \left[ \frac{-243}{5} + \frac{7 \cdot 27}{3} + 3 \cdot 9 \right]
   \]
   Simplify:
   \[
   V = \pi \left[ \frac{-243}{5} + 63 + 27 \right]
   \]
   Combine terms:
   \[
   V = \pi \left[ \frac{-243}{5} + 90 \right]
   \]
   Convert 90 to a fraction with denominator 5:
   \[
   V = \pi \left[ \frac{-243}{5} + \frac{450}{5} \right]
   \]
   \[
   V = \pi \left[ \frac{207}{5} \right]
   \]
   Therefore, the volume is:
   \[
   V = \frac{207\pi}{5} \text{ cubic units}.
   \]

Let me know if you need any further clarification or a breakdown of a specific part of the solution.

### Follow-up Questions:
1. What is the washer method and when is it used in volume problems?
2. How would the problem change if the region was rotated around the x-axis instead of \( y = -1 \)?
3. Could this problem be solved using the cylindrical shell method, and how would that setup differ?
4. How does shifting the axis of rotation affect the volume integral in general?
5. What are the steps for finding intersection points when rotating bounded regions?

### Tip: 
When rotating around a line that is not one of the coordinate axes, remember to adjust the functions accordingly by finding the distance from each curve to the axis of rotation.

Find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 3x about the line y = -1.

Find the volume of the solid obtained by rotating the region bounded by the curves y = x^2 and y = 3x about the line y = -1. This problem involves using the washer method and definite integrals to calculate the volume of a solid of revolution.

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